# Question #b6028

Jan 5, 2018

$3 \ln | \frac{x - 2}{x - 1} | + C$

#### Explanation:

We have that:

$\int \frac{3}{\left(x - 1\right) \left(x - 2\right)} \mathrm{dx}$

the best way to proceed would be by partial fractions; so we should split the quantity to be integrated like so:

$\frac{3}{\left(x - 1\right) \left(x - 2\right)} = \frac{A}{x - 1} + \frac{B}{x - 2}$

Now multiplying this equation through by the denominator $\left(x - 1\right) \left(x - 2\right)$:

$\to 3 = A \left(x - 2\right) + B \left(x - 1\right)$

Let $x = 2$ to cancel the A term:

$\to 3 = B$

And also let $x = 1$ to cancel the B term:

$3 = - A \to A = - 3$ so we now have values for $A$ and $B$ and so it follows that:

$\int \frac{3}{\left(x - 1\right) \left(x - 2\right)} \mathrm{dx} = \int \frac{3}{x - 2} - \frac{3}{x - 1} \mathrm{dx}$

The integral can now be evaluated straightforwardly as:

$= 3 \ln | x - 2 | - 3 \ln | x - 1 | + C = 3 \ln | \frac{x - 2}{x - 1} | + C$